# How do you write #f(x) = 3x^2 + 6x + 1# in vertex form?

Complete the square:

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To write ( f(x) = 3x^2 + 6x + 1 ) in vertex form, first factor out the coefficient of ( x^2 ) from the ( x^2 ) and ( x ) terms, completing the square. This gives ( f(x) = 3(x^2 + 2x) + 1 ).

Now, to complete the square, add and subtract ((2/2)^2 = 1) inside the parentheses, but since you added 1 inside the parentheses, you need to subtract 3 as well outside to keep the expression equivalent. This gives:

( f(x) = 3(x^2 + 2x + 1 - 1) + 1 )

( f(x) = 3((x + 1)^2 - 1) + 1 )

( f(x) = 3(x + 1)^2 - 3 + 1 )

( f(x) = 3(x + 1)^2 - 2 )

Therefore, the vertex form of ( f(x) = 3x^2 + 6x + 1 ) is ( f(x) = 3(x + 1)^2 - 2 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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